Uniform convergence on compact sets and on $\mathbb{R}$

Question

I am trying to determine where the series of functions $ \sum_{n=1}^\infty \frac{x \sin(n^2x)}{n^2}$ is uniformly convergent. I already determined that the convergence is uniform for $x \in S \subset \mathbb{R}$ where $S$ is compact. When $S$ is compact then $|x| \leq C$ for some constant $C$ and

$$\left|\frac{x \sin(n^2x)}{n^2}\right| = \frac{|x||\sin(n^2x)|}{n^2} \leq \frac{C}{n^2}$$

Since $\sum_{n=1}^\infty \frac{C}{n^2}$ converges the Weierstrass test implies uniform convergence. I also want to determine if the convergence is uniform on $x\in\mathbb{R}$ where $x$ can be unbounded.

It is not clear to me how to prove or disprove uniform convergence on the entire real line.

Answer 1

If the series converges uniformly on $\mathbb{R}$, then for any $\epsilon > 0$ there exists $N_0 \in \mathbb{N}$ such that for all $N \geqslant N_0, x \in \mathbb{R}$ we would have

$$\left|x\sum_{n=N}^\infty \frac{\sin(n^2x)}{n^2}\right|= \left|x\sum_{j=0}^\infty \frac{\sin(N+j)^2x)}{(N+j)^2}\right|< \epsilon,$$

and it would follow that $\displaystyle \left|\sum_{j=0}^\infty \frac{\sin((N+j)^2x_q)}{(N+j)^2}\right| < \frac{\epsilon}{x_q}$ for all $q \in \mathbb{N}$ and $x_q = \frac{\pi}{2} + 2q\pi$.

Note that $\sin((N+j)^2x_q) = \sin\left((N+j)^2\frac{\pi}{2}\right)$ which equals $0$ if both $N$ and $j$ are odd and equals $\sin\left(N^2 \frac{\pi}{2}\right) = \pm 1$ if $N$ is odd and $j = 2k$ is even.

Hence, for any odd $N \geqslant N_0$ and all $q \in \mathbb{N}$ we have

$$\frac{\epsilon}{x_q} > \left|\sum_{j=0}^\infty \frac{\sin((N+j)^2x_q)}{(N+j)^2}\right| = \left|\sin\left(N^2 \frac{\pi}{2}\right)\right| \sum_{k=0}^\infty \frac{1}{(N+2k)^2} = \sum_{k=0}^\infty \frac{1}{(N+2k)^2} $$

This is impossible as the LHS converges to $0$ as $q \to \infty$ and the series on the RHS converges to a positive value. Therefore, the convergence is not uniform on $\mathbb{R}$.

Answer 2

Assume that the series converges uniformly.

So exists $m \in \Bbb{N}$ such that $$A:=\sup_{x \in \Bbb{R}}\left|\sum_{n=m}^{\infty}\frac{x\sin{n^2x}}{n^2}\right|<1$$

But $A \geq \limsup_{x \to +\infty} |x| \left|\sum_{n=m}^{\infty}\frac{\sin{n^2x}}{n^2}\right|=+\infty$

since

$\left|\sum_{n=m}^{\infty}\frac{x\sin{n^2x}}{n^2}\right| \leq C,\forall x \in \Bbb{R}$ Apply the same argument for unbounded subsets of the real line.

Answer 3

Let $f_m(x)=\sum_{n=1}^m\frac {x\sin (n^2x)}{n^2}.$

If $(f_m)_{m\in \Bbb N}$ converged uniformly on $\Bbb R$ then $0=\lim_{m\to \infty}\|f_m-f_{m-1}\|_{\Bbb R}$ where $\|f_m-f_{m-1}\|_{\Bbb R}=\sup \{|f_m(x)-f_{m-1}(x)|:x\in \Bbb R\}.$

But with $x_m=m^2\pi +\frac {\pi}{2m^2}$ we have, for $m>1,$ that $$\|f_m-f_{m-1}\|_{\Bbb R}\ge |f_m(x_m)-f_{m-1}(x_m)|=$$ $$=\left| \frac {x_m\sin (m^2x_m)}{m^2}\right|=$$ $$=\frac {x_m}{m^2}>\frac {m^2\pi}{m^2}=\pi.$$